Set forms a basis, and span help

The last row shows that k_5 = 0. This means that p is a linear combination of the p_1 through p_4.In summary, the set S = {p1, p2, ..., p5} is a basis for the set of polynomials of degree 4 or less, and p belongs to the span of S.
  • #1
6
0

Homework Statement


http://img16.imageshack.us/img16/6606/50381320.jpg [Broken]


Homework Equations


Please see above picture


The Attempt at a Solution


I believe for question a) I just need to add up all the matrices and then row reduce to RREF, which gives me:
[1,0,0]
[0,1,0]

or Do I row reduce each matrices?

I'm really not sure how to do b), any tips would be great!

Thanks
 
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  • #2
think about linear combinations...

in a) can every N[tex]\in[/tex] M23 be written as a linear combination of the matricies given?

you could also think about what a simpler basis for M23 is and how they're related

similarly in b) can the equation be written as a linear combination of the basis equations?
 
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  • #3
I only wish to elaborate on Lanedance's post with a systematic spin.

billli said:

The Attempt at a Solution


I believe for question a) I just need to add up all the matrices and then row reduce to RREF, which gives me:
[1,0,0]
[0,1,0]

or Do I row reduce each matrices? Certainly not!
Think of the matrices in the set S as vectors in the vector space [tex]M_{23}[/tex] (the set of all 2x3 matrices). A basis for [tex]M_{23}[/tex] is
[tex]\mathbf{e_1} = \begin{pmatrix}1 & 0 & 0\\ 0&0&0\end{pmatrix},[/tex] [tex]\mathbf{e_2} = \begin{pmatrix}0 & 1 & 0\\ 0&0&0\end{pmatrix}, \ldots, [/tex] [tex]\mathbf{e_6} = \begin{pmatrix}0 & 0 & 0\\ 0&0&1\end{pmatrix}.[/tex]

The first vector in S can then be written as [tex]\mathbf{v_1} = [1 \; 2 \; 3 \; 4 \; 5 \; 6]^T[/tex] with respect to this basis, and similarly for the others, e.g., [tex]\mathbf{v_6} = [12 \; 10 \; 12 \; 12 \; 20 \; 18]^T[/tex].

So to systematically determine whether the vectors in S are linearly independent (and hence form a basis), i.e., to determine whether the equation [tex]k_1\mathbf{v_1} + \ldots + k_6\mathbf{v_6}=\mathbf{0}[/tex] has a nontrivial solution, one can row reduce the corresponding matrix [tex][\mathbf{v_1} \ldots \mathbf{v_6}][/tex].

billli said:
I'm really not sure how to do b), any tips would be great!
It suffices to show that S spans the set of all such (i.e., of degree 4 or less) polynomials, which amounts to determining whether p1, p2, ..., p5 are linearly independent. Taking e1 = 1, e2 = x, ..., e5 = x^4, this is just like the part (a).

-------------
Of course, in part (a) for example, if you can spot a non-trivial linear combination of the [tex]\mathbf{v_i}'s[/tex] that sums to [tex]\mathbf{0}[/tex], you are done. (I don't think it will be helpful if someone gives you one straight off the bat.)
 
  • #4
if you do look at the large matrix suggested by unco, its also worth considering the determinant and how it relates to linear independence and row reduction
 
  • #5
http://img239.imageshack.us/img239/6250/picture2.png [Broken]
Sorry, but I do not know how to use Tex. And thanks for the tips!

From this, a) does not form the basis, and for b) p(x) belongs to the span (not too sure about b) as it has a row of zeros)
 
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  • #6
billli said:
From this, a) does not form the basis,
Right,though the last column of zeros is superfluous.

billli said:
and for b) p(x) belongs to the span (not too sure about b) as it has a row of zeros)
This time taking vectors in row form (which is fine), the matrix you have here (albeit again with superfluous column of zeros) is what you would set up to see if {p, p1, p2, ..., p5} is linearly independent. You have shown they are not. But this does not necessarily mean p is a linear combination of p1, ... p5. For example, consider the set {(1,0), (2,0), (0,1)}: it is linearly dependent but (0,1) is not a linear combination of (1,0) and (2,0).

Rather, as said earlier, if you verify that S={p1, p2, ..., p5} is linearly independent (just as you did in part (a)), then this set forms a basis for the set of polynomials, so span(S) = the whole set and so certainly p belongs to span(S).
 
  • #7
Thanks for the reply Unco!
from matlab:
http://img156.imageshack.us/img156/582/set.jpg [Broken]
so P(x) belongs to the span of S. as it has a unique solution.
 
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  • #8
billli said:
...so P(x) belongs to the span of S. as it has a unique solution.
The first 5 columns of your row-reduced matrix show that S forms a basis. This is enough.

The last (sixth) column does give you the coefficients of the linear combination of p1, ..., p5 giving p.

The entire 5x6 matrix is the augmented matrix for the system

[tex][\mathbf{p_1} \, \ldots \, \mathbf{p_5}][k_1 \, \ldots \, k_5]^T = \mathbf{p},[/tex]

where [tex]k_i[/tex] is the coefficient of [tex]\mathbf{p_i}[/tex] in such a linear combination, and one wishes to solve for [tex][k_1 \, \ldots \, k_5]^T[/tex].

The point is that not all the coefficients of the linear combination, i.e., the k_i, are non-zero.
 

What is a basis in linear algebra?

A basis is a set of linearly independent vectors that can be used to represent any vector in a vector space. It is the smallest set of vectors that can span the entire space.

What is the significance of a basis?

A basis is important because it allows us to decompose any vector into a unique combination of basis vectors. This makes it easier to perform calculations and understand the properties of vectors and vector spaces.

How is a set related to a basis?

A set of vectors forms a basis if it is linearly independent and can span the entire vector space. This means that the set can be used as a basis to represent any vector in the space.

What is the difference between a basis and span?

A basis is a set of vectors, while span is the set of all possible linear combinations of those vectors. In other words, a basis forms the foundation for the span of a vector space.

Can a set that is not a basis still span a vector space?

Yes, a set of vectors can span a vector space without being a basis. However, a set that is not a basis may have redundant or unnecessary vectors, making it less efficient for performing calculations.

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